# Spectral decomposition of product of modular functions

The eigenfunctions of the Laplacian on $$SL(2,\mathbb Z)\backslash \mathbb H$$ are known to be given by three types: the constant function, the real analytic Eisenstein series (which come in a continuous family), and the Maass cusp forms (which come in a discrete family -- the first few given e.g. here). Let us denote $$E_s(\tau, \bar\tau)$$ as the real analytic Eisenstein series which scales as $$y^s + \frac{\Lambda(1-s)}{\Lambda(s)}y^{1-s}$$ as $$y \rightarrow \infty$$, where $$y = \text{Im}(\tau)$$, $$\Lambda(s) = \pi^{-s} \Gamma(s)\zeta(2s)$$; and $$\nu_i$$ denote the $$i^{\text{th}}$$ discrete cusp form.  My question is: Is there a simple expression for $$\int_{SL(2,\mathbb Z)\backslash \mathbb H} \frac{dxdy}{y^2} E_{s_1}(\tau, \bar\tau) E_{s_2}(\tau, \bar\tau) \overline{\nu_i(\tau,\bar\tau)}$$? (We can take $$s_i \in 1/2 + i \mathbb R$$.) For example, if there were only one $$E_s$$ in the previous expression, this would vanish due to orthonormality. What about when there is a product of two of them?

The motivation is the following --- it is known that any modular invariant function in $$L^2$$ can be expanded in Eisensteins on the principal series and Maass cusp forms. If we take the product of two modular invariants in $$L^2$$ each without support on the discrete cusp forms, does the product necessarily also have no support on the discrete series (let us assume the product remains in $$L^2$$, or at least has good decomposition..)?  Similarly are there simple or known expressions for:

$$\int_{SL(2,\mathbb Z)\backslash \mathbb H} \frac{dxdy}{y^2} E_{s_1}(\tau, \bar\tau) E_{s_2}(\tau, \bar\tau) \overline{E_{s_3}(\tau,\bar\tau)}$$

$$\int_{SL(2,\mathbb Z)\backslash \mathbb H} \frac{dxdy}{y^2} E_{s_1}(\tau, \bar\tau) \nu_{i_1}(\tau, \bar\tau) \overline{\nu_{i_2}(\tau,\bar\tau)}$$

$$\int_{SL(2,\mathbb Z)\backslash \mathbb H} \frac{dxdy}{y^2} \nu_{i_1}(\tau, \bar\tau) \nu_{i_2}(\tau, \bar\tau) \overline{\nu_{i_3}(\tau,\bar\tau)}$$ known in the literature?

(The first of these three may not converge but hopefully can be written with delta functions or something.) Thank you!

I'm guessing you're a physicist by your unusual notation! The real-analytic Eisenstein series is usually denoted $$E(z,s)$$ (where $$z = \tau$$ in your notation), not $$E_s(\tau,\bar{\tau})$$ (why does $$\bar{\tau}$$ appear?). I'll also let $$f(z)$$ denote a cusp form rather than $$\nu_i(\tau,\bar{\tau})$$ (I've never seen this notation used).

Anyway, the question asks (among other things) the following. Given two cusp forms $$f_1,f_2$$ and an Eisenstein series $$E(z,s)$$, determine a closed formula for $$\int_{\Gamma \backslash \mathbb{H}} f_1(z) f_2(z) E(z,s) \, d\mu(z).$$ Here $$\Gamma = \mathrm{SL}_2(\mathbb{Z})$$ and $$d\mu(z) = y^{-2} \, dx \, dy$$.

This question has a nice answer, as does the same question where one allows any triple product involving various cusp forms and Eisenstein series. When at least one form is an Eisenstein series, this is the Rankin-Selberg method; when all three are cusp forms, this is known as the Watson-Ichino triple product formula. In Section 2.2 of this paper of mine, I write out the exact formulae for these in several cases, and provide some references.

Let me give a more detailed explanation of this in the case written above. It is known, going back to work of Hecke, that any Maass cusp form $$f$$ can be written as a finite linear combination of Hecke-Maass cusp forms; these are eigenfunctions of the Hecke operators and have Fourier expansions of the form $$f(z) = \rho_f \sum_{n = 1}^{\infty} \frac{\lambda_f(n)}{\sqrt{n}} W_{0,it_f}(4\pi ny) (e^{2\pi i nx} + \epsilon_f e^{-2\pi i nx}).$$ Here $$\rho_f$$ is a normalising constant, $$\lambda_f(n)$$ is the $$n$$-th Hecke eigenvalue (which is multiplicative as an arithmetic function and conjecturally satisfies $$\lambda_f(p) \in [-2,2]$$ for any prime $$p$$), $$t_f$$ is the spectral parameter of $$f$$ (so that $$f$$ has Laplacian eigenvalue $$\lambda_f = 1/4 + t_f^2$$), $$W_{\alpha,\beta}$$ denotes the Whittaker function (in this case, it is basically just a $$K$$-Bessel function), and $$\epsilon_f \in \{1,-1\}$$ is the parity of $$f$$ (so that $$f(-\overline{z}) = \epsilon_f f(z)$$).

So we may assume that $$f_1$$ and $$f_2$$ are of this form. One then unfolds the integral by inserting the identity $$E(z,s) = \sum_{\gamma \in \Gamma_{\infty} \backslash \Gamma} \Im(\gamma z)^s,$$ using the fact that $$f_1(\gamma z) f_2(\gamma z) = f_1(z) f_2(z)$$, and that $$\Gamma_{\infty} \backslash \mathbb{H} \cong \{z : x \in [0,1], \ y > 0\}$$. One then inserts the Fourier expansions of $$f_1$$ and $$f_2$$ and evaluates the $$x$$ integral, then the $$y$$ integral; this latter integral is essentially the Mellin transform of the product of two $$K$$-Bessel functions, which has an explicit expression in terms of products of gamma functions.

All in all, one finds that the triple product is equal to $$\frac{\rho_{f_1} \rho_{f_2} \pi^{-s}}{\Gamma(s)} \prod_{\pm_1, \pm_2} \Gamma\left(\frac{s \pm_1 it_{f_1} \pm_2 it_{f_2}}{2}\right) \sum_{n = 1}^{\infty} \frac{\lambda_{f_1}(n) \lambda_{f_2}(n)}{n^s}.$$ (At least if $$\epsilon_{f_1} = \epsilon_{f_2}$$; otherwise, the gamma functions are changed slightly.) Here we are initially assuming that $$\Re(s) > 1$$; however, by analytic continuation, this extends meromorphically to all of $$\mathbb{C}$$ (with a simple pole at $$s = 1$$ if $$f_1 = f_2$$ and no poles otherwise). Let $$\zeta(s) = \sum_{n = 1}^{\infty} n^{-s}$$ denote the Riemann zeta function. Then we define the Rankin-Selberg $$L$$-function $$L(s,f_1 \times f_2)$$ to be $$L(s,f_1 \times f_2) = \zeta(2s) \sum_{n = 1}^{\infty} \frac{\lambda_{f_1}(n) \lambda_{f_2}(n)}{n^s}.$$ Again, this is initially defined for $$\Re(s) > 1$$ in this way, but extends meromorphically to $$\mathbb{C}$$. Moreover, this has a nice functional equation, which can be proven using the functional equation for $$E(z,s)$$.

Next, let me discuss the same problem for $$\int_{\Gamma \backslash \mathbb{H}} f(z) E\left(z,\frac{1}{2} + it\right) E(z,s) \, d\mu(z).$$ The key point is that $$E(z,1/2 + it)$$ has a nice Fourier expansion; $$t$$ replaces $$t_f$$, while $$\lambda_f(n)$$ is replaced by $$\lambda(n,t) = \sum_{ab = n} a^{it} b^{-it}$$. The exact same method works, except that the ensuing Rankin-Selberg $$L$$-function factorises as the product of two $$L$$-functions: $$\zeta(2s) \sum_{n = 1}^{\infty} \frac{\lambda_f(n) \lambda(n,t)}{n^s} = \sum_{m = 1}^{\infty} \frac{\lambda_f(m)}{m^{s + it}} \sum_{n = 1}^{\infty} \frac{\lambda_f(n)}{n^{s - it}} = L(s + it,f) L(s - it,f).$$

Now we consider $$\int_{\Gamma \backslash \mathbb{H}}^{\mathrm{reg}} E\left(z,\frac{1}{2} + it_1\right) E\left(z,\frac{1}{2} + it_2\right) E(z,s) \, d\mu(z).$$ I have written $$\int^{\mathrm{reg}}$$ to denote the fact that we must regularise (or renormalise) this integral in order for it to converge. This goes back to work of Zagier. Here not only does the Rankin-Selberg $$L$$-function factorise into the product of two $$L$$-functions, but each of these $$L$$-functions factorises further into the product of two copies of the Riemann zeta function.

Finally, let me consider $$\int_{\Gamma \backslash \mathbb{H}} f_1(z) f_2(z) f_3(z) \, d\mu(z).$$ Here the Rankin-Selberg method does not apply, since we cannot unfold as there is no Eisenstein series present. Nonetheless, this triple product can be shown (via methods coming from representation theory) to be associated to $$L$$-functions. This was done in Watson's thesis and was further expanded upon by Ichino. What they show is an identity relating the square of the absolute value of this integral to a certain $$L$$-function, namely $$L(1/2, f_1 \times f_2 \times f_3)$$.

• Great answer! If I might make a tiny quibble, I think the result you ascribe to Watson's thesis (2002) actually goes back to Garrett's 1987 Annals paper. Nov 19, 2020 at 8:01
• Sure. I think Garrett’s paper only deals with certain special cases, if memory serves. Nov 19, 2020 at 8:03
• Dear Peter, Thank you so much for the extremely detailed response! You are right, we are indeed physicists :). The $\bar{\tau}$ dependence I wrote is just to emphasize that the function is non-holomorphic but i suppose it is just notation. We actually had come across this unfolding” trick, but didn’t take it all the way to completion, and didn’t see the connection to L functions, so thank you very much for the answer! One quick followup question (in the next comment due to character limitation)--- Dec 4, 2020 at 21:00
• It is clear how to evaluate numerically the L function for Re(s)>1 (since it converges) and Re(s)<0 (from the functional equation). But for our application of interest, we need to evaluate the L function at Re(s)=1/2. Are there globally convergent expressions for these L functions that one can use to evaluate it numerically? I see that for the Riemann zeta function, there are some e.g. in en.wikipedia.org/wiki/…. Are there similar expressions for L functions? Best, Nathan Dec 4, 2020 at 21:00
• @nathanbenjamin The expressions you're looking for are known as the approximate functional equation; see, for example, Theorem 5.3 of Iwaniec-Kowalski. With that being said, actual numerical computations are pretty difficult. In analytic number theory, a key problem of interest is bounding these $L$-functions for $\Re(s) = 1/2$ in terms of the spectral parameters; there is an "easy" bound known as the convexity bound. Any power savings on this is a subconvex bound, and there is a plethora of literature on this. Dec 11, 2020 at 3:25